How do you differentiate #p = 2log_3(5^s) - log_3(4^s)#?

1 Answer
Jan 24, 2017

Answer:

#(dp)/(ds)=log_3(25/4)#

Explanation:

Note we can use the rule #log_a(b^c)=clog_a(b)#. Then:

#p=2slog_3(5)-slog_3(4)#

Note that #2log_3(5)# and #log_3(4)# are both constants, so the derivative is simply:

#(dp)/(ds)=2log_3(5)-log_3(4)#

If you wish, use #clog_a(b)=log_a(b^c)# and #log_m(n)-log_m(q)=log_m(n/q)#.

#(dp)/(ds)=log_3(25/4)#